Momentum momentum Momentum and Its Conservation - Lesson 1 - The Impulse-Momentum Change Theorem Momentum Real-World Applications The sports announcer says, "Going into the all-star break, the Chicago White Sox have the momentum." The headlines declare "Chicago Bulls Gaining Momentum." The coach pumps up his team at half-time, saying "You have the momentum; the critical need is that you use that momentum and bury them in this third quarter." Momentum is a commonly used term in sports. A team that has the momentum is on the move and is going to take some effort to stop. A team that has a lot of momentum is really on the move and is going to be hard to stop. Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team that is on the move has the momentum. If an object is in motion (on the move) then it has momentum. Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion. The amount of momentum that an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving.

Momentum depends upon the variables mass and velocity . In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object. Momentum = mass • velocity In physics, the symbol for the quantity momentum is the lower case p. Thus, the above equation can be rewritten as p = m • v where m is the mass and v is the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity. The units for momentum would be mass units times velocity units. The standard metric unit of momentum is the kg•m/s. While the kg•m/s is the standard metric unit of momentum, there are a variety of other units that are acceptable (though not conventional) units of momentum. Examples include kg•mi/hr, kg•km/hr, and g•cm/s. In each of these examples, a mass unit is multiplied by a velocity unit to provide a momentum unit. This is consistent with the equation for momentum. Momentum as a Vector Quantity Momentum is a vector quantity. As discussed in an earlier unit, a vector quantity is a quantity that is fully described by both magnitude and direction.

To fully describe the momentum of a 5-kg bowling ball moving westward at 2 m/s, you must include information about both the magnitude and the direction of the bowling ball. It is not enough to say that the ball has 10 kg•m/s of momentum; the momentum of the ball is not fully described until information about its direction is given. The direction of the momentum vector is the same as the direction of the velocity of the ball. In a previous unit, it was said that the direction of the velocity vector is the same as the direction that an object is moving. If the bowling ball is moving westward, then its momentum can be fully described by saying that it is 10 kg•m/s, westward. As a vector quantity, the momentum of an object is fully described by both magnitude and direction. The Momentum Equation as a Guide to Thinking From the definition of momentum, it becomes obvious that an object has a large momentum if both its mass and its velocity are large. Both variables are of equal importance in determining the momentum of an object. Consider a Mack truck and a roller skate moving down the street at the same speed. The considerably greater mass of the Mack truck gives it a considerably greater momentum.

Yet if the Mack truck were at rest, then the momentum of the least massive roller skate would be the greatest. The momentum of any object that is at rest is 0. Objects at rest do not have momentum - they do not have any " mass in motion ." Both variables - mass and velocity - are important in comparing the momentum of two objects. The momentum equation can help us to think about how a change in one of the two variables might affect the momentum of an object. Consider a 0.5-kg physics cart loaded with one 0.5-kg brick and moving with a speed of 2.0 m/s. The total mass of loaded cart is 1.0 kg and its momentum is 2.0 kg•m/s. If the cart was instead loaded with three 0.5-kg bricks, then the total mass of the loaded cart would be 2.0 kg and its momentum would be 4.0 kg•m/s. A doubling of the mass results in a doubling of the momentum.

Similarly, if the 2.0-kg cart had a velocity of 8.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 16.0 kg•m/s (instead of 4.0 kg•m/s). A quadrupling in velocity results in a quadrupling of the momentum. These two examples illustrate how the equation p = m•v serves as a "guide to thinking" and not merely a "plug-and-chug recipe for algebraic problem-solving." Check Your Understanding Express your understanding of the concept and mathematics of momentum by answering the following questions. Click the button to view the answers. 1. Determine the momentum of a ... a. 60-kg halfback moving eastward at 9 m/s. b. 1000-kg car moving northward at 20 m/s. c. 40-kg freshman moving southward at 2 m/s. A. p = 40 000 units (doubling the velocity will double the momentum) B. p = 60 000 units (tripling the velocity will triple the momentum) C.

p = 40 000 units (doubling the mass will double the momentum) D. p = 80 000 units (doubling the velocity will double the momentum and doubling the mass will also double the momentum; the combined result is that the momentum is doubled twice -quadrupled) 3. A halfback (m = 60 kg), a tight end (m = 90 kg), and a lineman (m = 120 kg) are running down the football field. Consider their ticker tape patterns below. Compare the velocities of these three players. How many times greater are the velocity of the halfback and the velocity of the tight end than the velocity of the lineman? See Answer A. The tight end travels twice the distance of the lineman in the same amount of time. Thus, the tight end is twice as fast (vtight end = 6 m/s). The halfback travels three times the distance of the lineman in the same amount of time. Thus, the halfback is three times as fast (vhalfback = 9 m/s). B. Both the halfback and the tight end have the greatest momentum.

The each have the same amount of momentum - 540 kg*m/s. The lineman only has 360 kg*m/s. Momentum | Define Momentum at Dictionary.com momentum [moh-men-tuh] /moʊˈmɛn tə/ (Show IPA), momentums. 1. force or speed of movement; impetus, as of a physical object or course of events: The car gained momentum going downhill. Her career lost momentum after two unsuccessful films. 2. Also called linear momentum. Mechanics. a quantity expressing the motion of a body or system, equal to the product of the mass of a body and its velocity, and for a system equal to the vector sum of the products of mass and velocity of each particle in the system. 3. 1690-1700; < Latin mōmentum; see moment Can be confused Examples from the Web for momentum Expand Contemporary Examples The budding relationship between the ornery sexagenarian and the precocious young child ultimately gives the film its momentum. BDS Loses by Losing Jon Haber July 9, 2012 Obama on points, but not the decisive knockout that the president needed to turn the momentum from the first debate.

Historical Examples When within a few rods of each other we ceased paddling, and drifted by with the momentum. The Forest Stewart Edward White We have also to define the moment of momentum of the planets around their axes. British Dictionary definitions for momentum Expand noun (pl) -ta (-tə), -tums 1. (physics) the product of a body's mass and its velocity p See also angular momentum 2. the impetus of a body resulting from its motion 3. C17: from Latin: movement; see moment Collins English Dictionary - Complete & Unabridged 2012 Digital Edition © William Collins Sons & Co. Ltd. 1979, 1986 © HarperCollins Publishers 1998, 2000, 2003, 2005, 2006, 2007, 2009, 2012 Word Origin and History for momentum Expand n. 1690s, scientific use in mechanics, "quantity of motion of a moving body," from Latin momentum "movement, moving power" (see moment ). Figurative use dates from 1782.

Online Etymology Dictionary, © 2010 Douglas Harper (mō-měn'təm) Plural momenta or momentums A vector quantity that expresses the relation of the velocity of a body, wave, field, or other physical system, to its energy. The direction of the momentum of a single object indicates the direction of its motion. Momentum is a conserved quantity (it remains constant unless acted upon by an outside force), and is related by Noether's theorem to translational invariance. In classical mechanics, momentum is defined as mass times velocity. The theory of Special Relativity uses the concept of relativistic mass. The momentum of photons, which are massless, is equal to their energy divided by the speed of light. In quantum mechanics, momentum more generally refers to a mathematical operator applied to the wave equation describing a physical system and corresponding to an observable; solutions to the equation using this operator provide the vector quantity traditionally called momentum. In all of these applications, momentum is sometimes called linear momentum. See also angular momentum , impulse. The American Heritage® Science Dictionary Copyright © 2002. Published by Houghton Mifflin. All rights reserved.

Expand momentum definition In physics , the property or tendency of a moving object to continue moving. For an object moving in a line , the momentum is the mass of the object multiplied by its velocity (linear momentum); thus, a slowly moving, very massive body and a rapidly moving, light body can have the same momentum. (See Newton's laws of motion .) Note: Figuratively, momentum can refer to the tendency of a person or group to repeat recent success: “The Bears definitely have momentum after scoring those last two touchdowns.” The American Heritage® New Dictionary of Cultural Literacy, Third Edition Copyright © 2005 by Houghton Mifflin Company. Published by Houghton Mifflin Company. All rights reserved. Momentum Momentum The momentum of a particle is defined as the product of its mass times its velocity. It is a vector quantity. The momentum of a system is the vector sum of the momenta of the objects which make up the system. If the system is an isolated system, then the momentum of the system is a constant of the motion and subject to the principle of conservation of momentum . The basic definition of momentum applies even at relativistic velocities but then the mass is taken to be the relativistic mass .

The most common symbol for momentum is p. The SI unit for momentum is kg m/s. You may insert numbers for any of the quantities. Then click on the text or symbol in the formula above for the quantity you wish to calculate.The numbers will not be forced to be consistent until you click on the desired quantity. Why is momentum represented by p? | Why You are here: Home Why is momentum represented by p? Why is momentum represented by p? Posted on May 18, 2011 by Hari M | 2 Comments Why is momentum represented by p? In physics there is a quantity that represents the product of mass and velocity of an object. That quantity is called as momentum. The popular equation for momentum according to the Newton’s law is that P=MV. Here, P stands for momentum, M stands for mass and V stands for Velocity of an object. If we observe the symbols used for various other quantities in physics, mass, time, velocity, force, acceleration are represented by their first letters. In this context, we might think that probably momentum is represented by Ã¢â‚¬ËœP’ as Ã¢â‚¬Ëœm’ was already used for mass.

If Ã¢â‚¬Ëœm’ was not used for mass, probably it could have been used for momentum. These usages existed since few decades. Before the Newton’s period it was believed that an object moves when a force is applied on it continuously. Later, Newton discovered Ã¢â‚¬Ëœmomentum’ which showed that object moves continuously without any force applied on it. The capacity or characteristic feature of the object to move continuously was called as impetus. Though Newton talked about momentum, he did not formulate the equation P=MV. This equation was formulated after some time. This study was later called as Newtonian mechanics. In the place of previous Ã¢â‚¬Ëœimpetus’, the name momentum was used by the people. Probably, Ã¢â‚¬ËœP’ was convenient symbol as per its previous name and that might have led to the representation of momentum with Ã¢â‚¬ËœP’. The letter is written in bold as the quantity is a vector.

Some interpret the reason for keeping PÃ¢â‚¬Â as symbol for momentum as the letter being derived from Latin word PetereÃ¢â‚¬Â which means to go and seekÃ¢â‚¬Â. This meaning is also related to the word Ã¢â‚¬Ëœimpetus’ which is the earlier term used for Ã¢â‚¬Ëœmomentum’. The word Ã¢â‚¬Ëœimpetus’ gives the meaning as to go and rush uponÃ¢â‚¬Â which was explained by the term PetereÃ¢â‚¬Â. Hence, momentum is represented by the letter PÃ¢â‚¬Â. Related Posts Impulse & Momentum - Summary – The Physics Hypertextbook Impulse & Momentum is a quantity that describes an object's resistance to stopping (a kind of "moving inertia"). is represented by the symbol p (boldface). is the product of an object's mass and velocity. p = mv is a vector quantity (since velocity is a vector and mass is a scalar). Impulse… is a quantity that describes the effect of a net force acting on an object (a kind of "moving force").

is represented by the symbol J (boldface). is the product of the average net force acting on an object and its duration. J = F̅Δt is the force–time integral. J = is a vector quantity (since force is a vector and time is a scalar). Impulse-Momentum Theorem The impulse-momentum theorem states that the change in momentum of an object equals the impulse applied to it. J = Δp If mass is constant, then… F̅Δt = mΔv If mass is changing, then… F dt = m dv + v dm The impulse-momentum theorem is logically equivalent to Newton's second law of motion (the force law). Units The SI unit of impulse is the newton second. The SI unit of momentum is the kilogram meter per second. These units of impulse and momentum are equivalent. [N s = kg m/s] is a measure of the efficacy of rocket propellants. if defined as impulse per mass (or thrust per mass flow rate) is equal to exhaust velocity and has the SI unit of meter per second.

Jsp[m/s] = I if defined as impulse per weight (or thrust per weight flow rate) is equal to exhaust velocity divided by acceleration due to gravity and has the SI unit of second. Jsp[s] = Momentum Notes You can explore more about this topic at the website for your textbook - click here and choose Chapter 7 for more links. Definitions Momentum is a quantity that can be thought of as a description of "moving inertia." The faster an object is moving, the more momentum it has. The more mass the moving object has, the more momentum it carries. Momentum is also a way to express how much force is needed to get something moving or to stop the object in a time of one second. Formally, the momentum of a moving object is defined as the product of the mass and velocity. Note that it is not just the speed, rather the velocity. Remember, the difference between the speed and the velocity is that velocity is a vector, it has a direction associated with it. This means that momentum is a vector quantity. It has both a magnitude (mass x speed) and a direction (the direction the object is moving).

The formula for momentum is: p = m v where p stands for momentum, m stands for mass (in kilograms), and v stands for velocity (in meters/sec). This means that momentum has units of kgm/s (kilogram-meter per second). Example: A runner has a mass of 75 kg and is running at 6 m/s (about 13.5 mph). The runner's momentum would be: p = m v p = (75 kg)x(6 m/s) p = 450 kgm/s What produces momentum in the first place? In order to move an object (give it some velocity) or to change the motion of a object (speed it up or slow it down), one must exert a force on the object. This force would act on the object for a certain length of time. The product of the force and the time the force acts is known as the impulse given to the object. This impulse produces a change in the object's momentum. The formula for Impulse is: I = F t where I stands for impulse, F stands for the force acting on the object, and t stands for the time the force acts on the object.

To give an object an impulse means that you are applying a force for a given period of time to change the object's motion or momentum. In fact, an impulse given to an object is equal to the change in momentum of the object. Start with the equation for Impulse: I = F t Substitute F = ma into the equation: I = (ma) t Substitute a=v/ t, where v is the change in velocity I = m( (change in v)/ t) t Note the time cancels: mv equals the change in momentum: I = change in p Examples: 1. A 1000 kg car is at rest initially. A 500 Newton force is applied to the car in the forward direction for 10 seconds. How fast is the car going at the end of the 10 seconds? Given: m=1000 kg ; F=500 N ; t=10 sec ; initial v = 0 m/s Want: final v Eqn: F t = m v ; (since initial v=0) v = (F t)/m = (500N x10s)/(1000 kg) = 5 m/s.

So the final speed of the car is 5 m/s. 2. A force of 1000 N acts to stop a 1000 kg car moving at 20 m/s. How much time does it take the car to stop? Given: F = -1000 N ; Change in v = -20m/s ; m=1000 kg Eqn: F t = m (change in v) t = m(change in v)/F = (1000kg)(-20m/s)/(-1000N) = 20 sec. So time required to stop the car is 20 seconds. Conservation of Momentum When two objects (or more) collide, the total momentum of the system (the combination of all the objects) remains the same, ie., momentum is conserved. Example #1: Collision between two identical masses, one initially at rest. We can investigate this using the following movie of a collision where two identical masses, one initially at rest, collide and bump off each other. Note the mass of each object is 2 kg. The initial velocity of Mass #1 is 1.0 m/s to the right.

This gives Mass #1 a momentum of (2kg)x(1.0 m/s) or 2 kgm/s to the right. The second mass isn't moving initially so it has zero momentum. The total momentum of the "system" (the two carts) is sum of the two momenta, namely 2 kgm/s to the right. After the collision, watch the momentum of Mass #1 decrease to zero and the momentum of Mass#2 increase to 2 kgm/s which is equal to the initial momentum of Mass #1. This means that the sum of the two momenta, that is, the total momentum of the system, remains the same at all times. To see a movie of this collision, click here. Example #2: Collision between two different masses, the less massive one initially at rest. Now we will look at a collision where the more massive #1 collides with a less massive #2 initially at rest. You will notice that the more massive #1 transfers some, but not all, of its initial momentum to Mass #2. Since there is some momentum left in the forward direction, Mass #1 continues to move forward, just at a slower speed.

Never the less, the total momentum of the system (the sum of the two momenta) remains constant. Play the movie and watch the momentum of each cart. Note the mass of #1 is 8 kg and the mass of #2 is 2 kg, which is 4 times less than Mass #1. The initial momentum of cart #1 is (8 kg)(1 m/s) or 8 kgm/s. The initial momentum of the second cart is zero. This means the total momentum before the collision is just 8 kgm/s. After the collision, cart #1 slows to 0.6 m/s, so its momentum is reduced to (8kg)(0.6 m/s) or just 4.8 kgm/s. The velocity of the second cart after the collision is 1.6 m/s, so its momentum is (2 kg)(1.6 m/s) or 3.2 kgm/s. Notice that the sum of the two carts' momenta is (4.8 + 3.2) kgm/s or 8 kgm/s. Hey, this is equal to the total momentum before the collision. Momentum is conserved again! Physics triumphs again.

To see a movie of this collision, click here. Example #3: Collision between two different masses, the more massive one initially at rest. This time we will have a lighter cart strike a more massive cart which is initially at rest. Surprising results are observed, play the movie by clicking here to see what happens . Here the moving cart #1 transfers more momentum that it had initially to Mass #2 (note the momentum of #2 after is 1.333 kgm/s while the initial momentum of #1 was only 1.0 kgm/s). Since #1 transferred more forward momentum than it had to start with to #2, the first cart must recoil (head backwards) with a momentum of 0.333 kgm/s. This backward momentum is designated with a negative sign in front of the momentum (-0.333 kgm/s). Thus the sum of the two momenta (-0.333 kgm/s + 1.333 kgm/s = 1.000 kgm/s) remains constant. Wow, this physics stuff is amazing. Ready for a practice test? Click here for a momentum/energy practice test .

StudyofMomentumandConservationofMomentum Study Of Momentum and Conservation of Momentum Momentum and Conservation of Momentum: Isaac Newton studied a phenomena referred to as momentum. He defined momentum as the quantity of the motion of a body that is described by the mathematical product of mass and velocity. In the MKS system (meter, kilogram, second) this would give momentum the units of kg m/s. Note: That though the units for momentum is similar to the units for force, 1 N = 1 kg m/s^2, it is not the same measurement at all. Momentum is simply kg m/s. Isaac Newton also explored the conservation of momentum which says that "the total momentum of a closed system before an event occurs is equal to the total momentum after the event occurs". To express this in another way, though the individual momentum of different objects may change within a closed system, the total momentum of the closed system remains the same. [Special Note: Conservation laws are found throughout the study of science. Physicists found that they were more easily able to study certain phenomena such as momentum using conservation laws, where, though individual quantities within a system may change, the overall sum total of a quantity remained constant.

The law of conservation of energy is another example of a conservation law.] Momentum: As defined by Newton, momentum is the product of an objectï¿½s mass and its velocity. The symbol for momentum is the lower case p. Momentum has to do with questions about both starting and stopping objects. Momentum is not just about the mass of an object or just about the velocity of an object, but it is about the combined effect of both measurements when an objectï¿½s motion changes. Consider this question, ï¿½ Which object has more momentum and would be harder to stop, a small sports car traveling with a high velocity or a large truck moving with a low velocity? ï¿½. Example: Calculate the momentum of a 10 kg object moving at 5 m/s. G: m = 10 kg, v = 5 m/s F: the momentum of the truck E: p = m v S: p = 10 kg * 5 m/s A: p = 50 kg m/s Impulse: When discussing the prospect of changing the momentum of an object, the term Impulse (symbol J*) is introduced.

Impulse is the product of the unbalanced (net) force applied to an object multiplied by the time interval over which the force is applied. The unit for impulse is the Ns, based on the fact that Impulse (J) equals the product of Force and the time interval over which the force is applied. (*Some sources use an upper case I, but on this page we use the upper case J, which is a very commonly used symbol for Impulse) An impulse causes a change in motion and is equal to the change in momentum that an object experiences because of the impulse. Calculating Impulse Example: Calculate the impulse experienced by a 7.5 kg object when a 20 N force is applied to it for 10 seconds. G: m = 7.5 kg, F = 20 N, t = 10 s F: the impulse experienced by the object E: Impulse = J = F t S: Impulse = J = 20 N * 10 s A: Impulse = J = 200 Ns Impulse and change in momentum: Impulse is the means by which momentum is changed, so we may say that Impulse equals change in momentum.

[rem: The phrase ï¿½change inï¿½ is symbolized by /\ (delta) symbol.] J = /\ p or F /\t = m /\ v NOTE: Though it is possible to change momentum by changing the mass of an object, most often we study change in momentum as the result of a change in an objectï¿½s velocity. Using the Impulse-momentum Theorem Example: A force of 20 N is applied to a 50 kg object over 10 seconds. Determine the objectï¿½s change in velocity. G: F = 20 N, m = 50 kg, t = 10 s F: the change in velocity of the object E: J = /\p, which is F t = m /\ v S: 20 N * 10 s = 50 kg * /\ v A: /\ v = 4 m/s In the above problem, if the v(initial) was 12 m/s, what would the v(final) equal?

G: v(initial) = 12 m/s, /\ v = 4 m/s F: the final velocity S: v(final) = 12 m/s + 4 m/s A: v(final) = 16 m/s Newtonï¿½s Third Law: During the study of forces we discussed Newtonï¿½s three laws of motion, but most of our attention was on his first and second laws. We are now going to turn our attention towards his third law. Authors often use the phrase ï¿½the law of action and reactionï¿½ when describing Newtonï¿½s third law. This statement describes the forces exerted by two different objects when they interact with one another. A classical example described in many physics texts is that of a cannon launching (firing) a cannon ball. When launched the cannon ball flies forward with a specific amount of momentum. At the same time the cannon experiences a recoil velecity, which results in it moving backwards with an equal amount of momentum. The cannon ball and the cannon both experience the same amount of momentum, only it is in opposite directions. The cannon ball's momentum is in the forward direction, and the cannon's momentum is in the reverse direction.

Another example is when you push on a wall the ï¿½wall pushes backï¿½. "Granted that the wall is not ï¿½making an effort to push back activelyï¿½, but it is by the very design and choice of materials that were used to construct it that it does resist your effort, thus, in a sense, it pushes back." In such a case we might say that its resistance represents a force pushing back. That "resistance force" is due to the very nature of the wall based upon the architects' design and the contractors' efforts in building the wall. Likewise, the stools, chairs, and the tables or benches used in your science classroom are designed to push back against your weight or the weight of any other object placed on them. Weight is a downward force and the force they supply as they support that downward force is an equal but opposite force that prevents the stool, chair, table or bench from collapsing. In the study of forces this ï¿½upward forceï¿½ that opposes gravity and is perpendicular to the surface upon which an object resides is called the normal force. Example: Consider two objects such as a book and a table. The book is resting on the surface of the table.

It is pressing down with a force, the force of gravity, on the top of the table. By the very nature of the design of the table and the choice of materials it is made from, it resists that downwards force and exerts an upward force equal but opposite in direction, meaning upwards in this case, to that of the weight of the book. An individual or single force cannot exist. For one object to produce a force on a second object the second object must produce a force on the first object. When baseballs and tennis balls are stuck, they both experience a force as the result of an impulse during contact. But these forces are not isolated, single forces. As they are being struck, the baseball bat and the tennis racquet are also experiencing equal but opposite forces, respectively, as the result of an impulse during contact. This can be described by the statement, "The momentum gained by one object, when it makes contact with a second object, is lost by the other object." Law of Conservation of Momentum: When studying conservation laws, such as the law of conservation of momentum, the term isolated system is used.

An "isolated system" is a system that only experiences internal changes, but experiences no changes as a whole due to external net (unbalanced) forces from without. A system can be very complex involving a large number of objects in a very large space. It can also be simply two object interacting with one another without anything else interfering with them. It is in this context that we will look at the conservation of momentum. The Law of Conservation of Momentum says, ï¿½The total momentum of an isolated system does not change.ï¿½ What this means is that the initial total momentum of an isolated system before an internal event like a collision between two objects takes place is equal to the final total momentum of that same isolated system after the event. The total momentum of an isolated system always remains the same. In a simple system, having only two objects, the momentum that one object loses in a collision is gained by the other object. Collisions: There are two categories of simple two object collisions. One is the elastic collision and the other is the inelastic collision. Both are explained below. Elastic Collision: In this situation one object collides with a second object after which each object moves off separate from the other.

At the moment of collision momentum is transferred between the two objects in such a way that the total momentum of the two together is conserved. In such a collision we expect velocity to change for each object, but not the mass. An example of such a collision is seen in both the game of billiards and the game of pool. Momentum is transferred form one ball to another ball during a collision. Though many automobile collisions involve only two objects the circumstances of crumple zones and the likelihood of pieces of both cars breaking off make the situation more complex. Elastic Collision Example: Object 1 (mass is 6 kg), moving along with a velocity of 10 m/s, collides with Object 2 (mass is 3 kg), moving along with a velocity of 5 m/s. If object 1ï¿½s velocity is decreased to 4 m/s by the collision, what will object 2ï¿½s velocity be after the collision? G: Before collision: m(1) = 6 kg, v(1) = 10 m/s, m(2) = 3 kg, v(2) = 5 m/s After the collision: the masses are the same, but v(1)!

= 4 m/s and v(2)! = ? [Note: the ! symbol means a new value for the same variable] F: the second object's velocity after the collision E: p(1)Before + p(2)Before = p(1)After + p(2)After . and so m(1) * v(1) + m(2) * v(2) = m(1) * v(1)! + m(2) * v(2)! S: Step1 6 kg * 10 m/s + 3 kg * 5 m/s = 6 kg * 4 m/s + 3 kg * v(2)! To simplify this long equationï¿½s substitution, look at the following steps: Step2 60 kg m/s + 15 kg m/s = 24 kg m/s + 3 kg * v(2)! Step3 75 kg m/s = 24 kg m/s + 3 kg * v(2)! Step4 75 kg m/s ï¿½ 24 kg m/s = 3 kg * v(2)!

Step5 51 kg m/s = 3 kg * v(2)! Step6 51 kg m/s / 3 kg = v(2)! A: v(2)! = 17 m/s Inelastic Collision: In a two object collision of this type, the two objects remain stuck together after the collision and move along as though they were a single object having a mass larger than either object"s individual mass. An example of this type of collision is seen in railroad switching yards, where one railcar rolls down a hill onto a siding where another railcar is sitting. When the rolling car strikes the stationary car the couplers lock together and the two cars roll together as one until friction slows them to a stop or someone applies a brake or a retarder to their wheels. Another example of an inelastic collision is a dart being shot at a small stationary block of wood. If the mass of the block of wood is sufficiently small the dart and the block of wood will move as one single object until stopped by friction or some other force. Inelastic Collision Example: A metal 0.50 kg glider carrying a 0.25 kg load on a frictionless air track is moving at 2.5 m/s.

This glider encounters a second identical glider carrying no load which is stationary. After the two gliders collide they stick together because of magnets attached to the ends of the gliders. What will the final velocity of the two equal as they continue moving while attached with one another? G: m(1) = 0.5 kg, m(1)(extra load) = 0.25 kg, v(1) = 2.5 m/s, m(2) = 0.50 kg, v(2) = 0 m/s F: final velocity of two gliders attached together E: p(1)Before + p(2)Before = p(1)After + p(2)After and so m(1) * v(1) + m(2) * v(2) = m(1) * v(1)! + m(2) * v(2)! but since v(1)! = v(2)!, the equation can be rewritten as m(1) * v(1) + m(2) * v(2) = ( (m(1) + m(2) ) * v! [where v!

is the combined velocity] S: (0.5 + 0.25) kg * 2.5 m/s + 0.5 kg * 0 m/s = (0.75 kg + 0.5 kg) * v! The simplification steps are as follows Step1 0.75 kg * 2.5 m/s + 0.5 kg * 0 m/s = 1.25 kg * v! Step2 1.875 kg m/s + 0 kg m/s = 1.25 kg * v! Step3 1.875 kg m/s = 1.25 kg * v! Step4 1.875 kg m/s / 1.25 kg = v! Step5 1.5 m/s = v! A: v! = 1.5 m/s Action ï¿½ Reaction Problems: Often in the study of physics we encounter a situation where two objects at rest are suddenly flung in opposite directions because of some force acting that separates them. The mechanism producing the motion in opposite directions can be something as simple as a compressed spring or something as complex as gunpowder or rocket fuel.

Two people on skates facing each other on very smooth ice could also be a model of this type of situation. One or both of them can push themselves apart. Also, a skater can move, if they throw something away from themselves. There is an action reaction situation. Likewise, people wonder how rockets get launched and fly through the air into and through space beyond the earth's atmosphere. They think that the rocket exhaust pushes against something, but it doesnï¿½t. Most often there is nothing to push against in the first place. A rocket moves by the process of pushing out hot exhaust gasses in the direction opposite to the one they want to go. This results in the rocket going in the opposite direction from the exhaust gas. The conservation of momentum tells us that the product of the mass and velocity of the hot exhaust gas going in one direction is equal to the product of the mass and velocity of the rocket going in the opposite direction. An untethered astronaut floating in space could move back towards the space craft by throwing a tool in the direction opposite to the space craft. Action - Reaction Example: A 500 kg cannon fires a 1.5 kg cannon ball through the air.

The cannon ball leaves the muzzle of the barrel of the cannon with a velocity of 25 m/s. What is the recoil velocity of the cannon? G: Before: m(1)(cannon) = 500 kg, v(1) = 0 m/s, m(2)(cannon ball) = 1.5 kg, v(2) = 0 m/s. [rem: before the cannon is fired the objects are stationary] After: v(1)! = ?, v(2) = 25 m/s F: the recoil velocity of the cannon E: p(1)Before + p(2)Before = p(1)After + p(2)After and so E: m(1) * v(1) + m(2) * v(2) = m(1) * v(1)! + m(2) * v(2)! but v(1) = 0 m/s and v(2) = 0 m/s whcih makes the left side of the equation equal to 0 kg m/s and thus 0 kg m/s = m(1) * v(1)! + m(2) * v(2)!

S: 500 kg * 0 m/s + 1.5 kg * 0 m/s = 500 kg * v(1)! + 1.5 kg * 25 m/s Simplification steps Step 1 0 kg m/s + 0 kg m/s = 500 kg * v(1)! + 37.5 kg m/s Step 2 0 kg m/s ï¿½ 37.5 kg m/s = 500 kg v(1)! Step 3 - 37.5 kg m/s / 500 kg = v(1)! Step 4 - 0.075 m/s = v(1)! A: v(1)! = - 0.075 m/s Momentum and Impulse Connection Momentum and Its Conservation - Lesson 1 - The Impulse-Momentum Change Theorem Momentum and Impulse Connection Real-World Applications As mentioned in the previous part of this lesson , momentum is a commonly used term in sports. When a sports announcer says that a team has the momentum they mean that the team is really on the move and is going to be hard to stop.

The term momentum is a physics concept. Any object with momentum is going to be hard to stop. To stop such an object, it is necessary to apply a force against its motion for a given period of time. The more momentum that an object has, the harder that it is to stop. Thus, it would require a greater amount of force or a longer amount of time or both to bring such an object to a halt. As the force acts upon the object for a given amount of time, the object's velocity is changed; and hence, the object's momentum is changed. The concepts in the above paragraph should not seem like abstract information to you. You have observed this a number of times if you have watched the sport of football. In football, the defensive players apply a force for a given amount of time to stop the momentum of the offensive player who has the ball. You have also experienced this a multitude of times while driving. As you bring your car to a halt when approaching a stop sign or stoplight, the brakes serve to apply a force to the car for a given amount of time to change the car's momentum. An object with momentum can be stopped if a force is applied against it for a given amount of time.

A force acting for a given amount of time will change an object's momentum. Put another way, an unbalanced force always accelerates an object - either speeding it up or slowing it down . If the force acts opposite the object's motion, it slows the object down. If a force acts in the same direction as the object's motion, then the force speeds the object up. Either way, a force will change the velocity of an object. And if the velocity of the object is changed, then the momentum of the object is changed. Impulse These concepts are merely an outgrowth of Newton's second law as discussed in an earlier unit. Newton's second law (Fnet = m • a) stated that the acceleration of an object is directly proportional to the net force acting upon the object and inversely proportional to the mass of the object. When combined with the definition of acceleration (a = change in velocity / time), the following equalities result. F = m • a F = m • ∆v / t If both sides of the above equation are multiplied by the quantity t, a new equation results. F • t = m • ∆v This equation represents one of two primary principles to be used in the analysis of collisions during this unit.

To truly understand the equation, it is important to understand its meaning in words. In words, it could be said that the force times the time equals the mass times the change in velocity. In physics, the quantity Force • time is known as impulse. And since the quantity m•v is the momentum, the quantity m•Δv must be the change in momentum. The equation really says that the Impulse = Change in momentum One focus of this unit is to understand the physics of collisions. The physics of collisions are governed by the laws of momentum; and the first law that we discuss in this unit is expressed in the above equation. The equation is known as the impulse-momentum change equation. The law can be expressed this way: In a collision, an object experiences a force for a specific amount of time that results in a change in momentum. The result of the force acting for the given amount of time is that the object's mass either speeds up or slows down (or changes direction). The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum.

Consider a football halfback running down the football field and encountering a collision with a defensive back. The collision would change the halfback's speed and thus his momentum. If the motion was represented by a ticker tape diagram , it might appear as follows: At approximately the tenth dot on the diagram, the collision occurs and lasts for a certain amount of time; in terms of dots, the collision lasts for a time equivalent to approximately nine dots. In the halfback-defensive back collision, the halfback experiences a force that lasts for a certain amount of time to change his momentum. Since the collision causes the rightward-moving halfback to slow down, the force on the halfback must have been directed leftward. If the halfback experienced a force of 800 N for 0.9 seconds, then we could say that the impulse was 720 N•s. This impulse would cause a momentum change of 720 kg•m/s. In a collision, the impulse experienced by an object is always equal to the momentum change. Representing a Rebounding Collision Now consider a collision of a tennis ball with a wall.

Depending on the physical properties of the ball and wall, the speed at which the ball rebounds from the wall upon colliding with it will vary. The diagrams below depict the changes in velocity of the same ball. For each representation (vector diagram, velocity-time graph, and ticker tape pattern), indicate which case (A or B) has the greatest change in velocity, greatest acceleration , greatest momentum change, and greatest impulse . Support each answer. Click the button to check your answer. See Answer a. The velocity change is greatest in case B. In each case the initial velocity is the same. In case B, the object rebounds in the opposite direction with a greater speed than in case A. This is equivalent to a change from +10 m/s to -5 m/s; whereas, case A has a change from +10 m/s to -2 m/s. b. The acceleration is greatest in case B. Acceleration depends on velocity change and the velocity change is greatest in case B (as stated above) c. The momentum change is greatest in case B. Momentum change depends on velocity change and the velocity change is greatest in case B (as stated above). d. The impulse is greatest in case B.

Impulse equals momentum change and the momentum change is greatest in case B (as stated above) Observe that each of the collisions above involve the rebound of a ball off a wall. Observe that the greater the rebound effect, the greater the acceleration, momentum change, and impulse. A rebound is a special type of collision involving a direction change in addition to a speed change. The result of the direction change is a large velocity change. On occasions in a rebound collision, an object will maintain the same or nearly the same speed as it had before the collision. Collisions in which objects rebound with the same speed (and thus, the same momentum and kinetic energy) as they had prior to the collision are known as elastic collisions. In general, elastic collisions are characterized by a large velocity change, a large momentum change, a large impulse, and a large force. Use the impulse-momentum change principle to fill in the blanks in the following rows of the table. As you do, keep these three major truths in mind: The impulse experienced by an object is the force•time. The momentum change of an object is the mass•velocity change. The impulse equals the momentum change. Click the button to view answers.

-4 m/s There are a few observations that can be made in the above table that relate to the computational nature of the impulse-momentum change theorem. First, observe that the answers in the table above reveal that the third and fourth columns are always equal; that is, the impulse is always equal to the momentum change. Observe also that if any two of the first three columns are known, then the remaining column can be computed. This is true because the impulse=force • time. Knowing two of these three quantities allows us to compute the third quantity. And finally, observe that knowing any two of the last three columns allows us to compute the remaining column. This is true since momentum change = mass • velocity change. There are also a few observations that can be made that relate to the qualitative nature of the impulse-momentum change theorem. An examination of rows 1 and 2 show that force and time are inversely proportional; for the same mass and velocity change, a tenfold increase in the time of impact corresponds to a tenfold decrease in the force of impact.

An examination of rows 1 and 3 show that mass and force are directly proportional; for the same time and velocity change, a fivefold increase in the mass corresponds to a fivefold increase in the force required to stop that mass. Finally, an examination of rows 3 and 4 illustrate that mass and velocity change are inversely proportional; for the same force and time, a twofold decrease in the mass corresponds to a twofold increase in the velocity change. We Would Like to Suggest ... Sometimes it isn't enough to just read about it. You have to interact with it! And that's exactly what you do when you use one of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of our Egg Drop Interactive . You can find it in the Physics Interactives section of our website. The Egg Drop Interactive immerses a learner into a Virtual Egg Drop activity in order to explore the effect of drop height, egg mass, and landing surface upon the outcome of the egg. See Answer The momentum change is the same for each cart. Momentum change equals the impulse; if each cart has the same impulse, then it would follow that they have the same momentum change.

2. In a physics demonstration, two identical balloons (A and B) are propelled across the room on horizontal guide wires. The motion diagrams (depicting the relative position of the balloons at time intervals of 0.05 seconds) for these two balloons are shown below. See Answer Balloon B has the greatest acceleration. The rate at which the velocity changes is greatest for Balloon B; this is shown by the fact that the speed (distance/time) changes most rapidly. See Answer Balloon B has the greatest impulse. Impulse is equal to momentum change. If balloon B has the greatest momentum change, then it must also have the greatest impulse. 3. Two cars of equal mass are traveling down Lake Avenue with equal velocities. They both come to a stop over different lengths of time. The ticker tape patterns for each car are shown on the diagram below. At what approximate location on the diagram (in terms of dots) does each car begin to experience the impulse? See Answer The collision occurs at approximately the ninth dot (plus or minus a dot). The diagram shows that it is at that location that the cars begin to slow down. See Answer Car A has the greatest acceleration. The velocity change of each car is the same.

(They start with the same velocity and each finish with zero velocity.) Yet car A accomplishes this change in less time. Car A accelerates "most rapidly." See Answer The impulse is the same for each car. The impulse equals the momentum change. If the momentum change is the same for each car, then so must be the impulse. 4. The diagram to the right depicts the before- and after-collision speeds of a car that undergoes a head-on-collision with a wall. In Case A, the car bounces off the wall. In Case B, the car crumples up and sticks to the wall. a. In which case (A or B) is the change in velocity the greatest? Explain. See Answer Case A has the greatest velocity change. The velocity change is -9 m/s in case A and only -5 m/s in case B. See Answer The impulse is greatest for Car A. The impulse equals the momentum change. If the momentum change is greatest for Car A, then so must be the impulse. d. In which case (A or B) is the force that acts upon the car the greatest (assume contact times are the same in both cases)? Explain.

See Answer The impulse is greatest for Car A. The force is related to the impulse (I=F*t). The bigger impulse for Car A is attributed to the greater force upon Car A. Recall that the rebound effect is characterized by larger forces; car A is the car which rebounds. 5. Jennifer, who has a mass of 50.0 kg, is riding at 35.0 m/s in her red sports car when she must suddenly slam on the brakes to avoid hitting a deer crossing the road. She strikes the air bag, that brings her body to a stop in 0.500 s. What average force does the seat belt exert on her? Impulse = F*t = 80 N * 0.1 s Impulse = 8 N*s Note that not all the numbers are necessary for computing the impulse; don't "force" the value of mass into the computation. 7. If a 5-kg object experiences a 10-N force for a duration of 0.10-second, then what is the momentum change of the object? Momentum Change = 1.0 kg*m/s The momentum change = mass*velocity change.

But since velocity change is not known another strategy must be used to find the momentum change. The strategy involves first finding the impulse (F*t = 1.0 N*s). Since impulse = momentum change, the answer is 1.0 N*s.

Question: In Physics, momentum is the product of mass and which other quantity?
Answer: VELOCITY